2012-02-18 22:11 阅读 52


When I use str_replace with an array as second argument like in the much upvoted and accepted answer to this question, I get an array to string conversion notice. Why? Example:

$str = 'a b a ba a';
$numerals = range(1, 10);
$str = str_replace('a', $numerals, $str);

gives :

PHP Notice: Array to string conversion in php shell code on line 1

and the following output:

Array b Array bArray Array

instead of

1 b 2 b3 4

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    dongxiao9583 dongxiao9583 2012-02-18 22:27

    You are trying to replace one character ('a') with multiple characters (1,2,3,4,5,6,7,8,9,10). PHP can't figure that out and tries to convert this array to string. When you're using string as $search parameter you must also use string as $replace parameter. An array as $replace parameter can be used only when $search is also an array. Quoting from documentation:

    If search and replace are arrays, then str_replace() takes a value from each array and uses them to search and replace on subject. If replace has fewer values than search, then an empty string is used for the rest of replacement values. If search is an array and replace is a string, then this replacement string is used for every value of search. The converse would not make sense, though.

    Here's the code that will work:

    $str = 'a b a ba a';
    $count = 1;
    while(($letter_pos = strpos($str, 'a')) !== false) {
        $str = substr_replace($str, $count++, $letter_pos, 1);
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